## Rule of Thumb for RC Design

Fire resistance (hrs) | Minimum beam width (mm) | Minimum thickness of floors (mm) | Minimum wall thickness (p<0.4%) | Min wall thickness (0.4% | Min wall thickness (p>1%) |
---|---|---|---|---|---|

0.5 | 200 | 75 | 150 | 100 | 75 |

1 | 200 | 95 | 150 | 120 | 75 |

1.5 | 200 | 110 | 175 | 140 | 100 |

2 | 200 | 125 | - | 160 | 100 |

3 | 240 | 150 | - | 200 | 150 |

4 | 280 | 170 | - | 240 | 180 |

Source: Concrete Centre

**Reinforcement for beams**

**Spacing**

- Maximum size of coarse aggregate plus 5mm

2. Bar size(whichever is greater)

**Number of bars**

**Size of bars**

Beam width (mm) | Bar diameter (assuming 35mm cover) | ||
---|---|---|---|

25 | 32 | 40 | |

300 | 3 | 3 | 2 |

350 | 4 | 3 | 3 |

400 | 5 | 4 | 3 |

450 | 6 | 5 | 4 |

500 | 7 | 5 | 4 |

550 | 8 | 6 | 5 |

600 | 9 | 7 | 6 |

650 | 10 | 8 | 6 |

700 | 11 | 9 | 7 |

750 | 12 | 10 | 8 |

800 | 13 | 10 | 8 |

900 | 15 | 12 | 9 |

1000 | 17 | 13 | 11 |

Source: Concrete Centre

**Areas of reinforcement**

Min percentages are shown in the table below, which is Table 3.25 of BS 8110

**Shear Links**

Bar diameter (mm) | 16 | 20 | 25 | 32 | 40 |
---|---|---|---|---|---|

Max spacing (mm) | 192 | 249240 | 300 | 384 | 480 |

Min link diameter (mm) | 6 | 6 | 8 | 8 | 10 |

Source: Concrete Centre

## Reinforced Concrete Design

Elastic response is due to applied loads but plastic behavior can be below and above yield stress.

Creep rate depends on the composition of the concrete and environmental conditions.

Similar to steel, concrete multi-storey buildings can either consist of portal frames or braced frames that depend on bracing or diaphragms with concrete core walls for lateral stability. However, for multi-storey buildings, lateral stability has several requirements:

Creep rate depends on the composition of the concrete and environmental conditions.

Similar to steel, concrete multi-storey buildings can either consist of portal frames or braced frames that depend on bracing or diaphragms with concrete core walls for lateral stability. However, for multi-storey buildings, lateral stability has several requirements:

- Stiff horizontal diaphragms must need to be used with core walls, such as constructing floors with insitu reinforced concrete. Concrete core walls (with a minimum thickness of 200mm for steel reinforcement placement and concreting) can be in the form of lift shafts or the surrounding walls of staircases.
- Bracing should be used throughout the height of the building, unless transfer structures are used.

## Reinforced Concrete: Column Design

**Scheme Design**

We will always design columns and other compression members where their vertical loads act concentrically to the neutral axis of the structural members. In these situations, these structural members are axially loaded by direct compressive stresses.

Concrete columns are structural members that help structural durability and resist and supports vertical loads. To distinguish concrete columns from concrete piers and walls, the bigger cross-section dimension should not be larger than four times its smaller dimension.

In practical applications, vertical loads act eccentrically to the neutral axis of the structural member. Therefore, in actual practice, both the compressive stresses that act concentrically to the neutral axis of the structural member AND the bending stresses induced by the compressive stresses that act eccentrically to the neutral axis of the structural member need to be accounted for in the structural design.

**We will only focus on compressive stresses that act concentrically to the neutral axis in scheme designs.**

Concrete columns are considered to be braced when the overall structure is designed to resist lateral loads. Braced columns are columns in a stability system with shear or core walls. Unbraced columns are columns in a system where the only structural elements supporting the overall stability of the structure are the columns.

Columns are short if slenderness is less than 15 for braced columns or 10 for unbraced columns.

- Short columns - Crushing failure is caused by direct compression stresses
- Slender columns - Lateral buckling and crushing failure are caused by direct compression stresses and bending stresses caused by eccentric compression stresses. The amount of failure depends on the end fixity conditions and the slenderness ratio, which is effective length divided by radius of gyration.

**1. Determine fy and fcu**

**2. Determine applied Live Load and Dead Load on the column**

**3. Determine tributary load area on the column**

**4. Determine the number of floors the column supports**

**5. Determine the total loads acting on the column by using the equation below**

Total Load, N = (LL + DL) x ULS Factor x # of Floors x Tributary Load Area x Elastic Shear Factor

where LL = Live Load

DL = Dead Load

ULS Factor = 1.6 (for conservative purposes)

Elastic Shear Factor = 1.25

**6. Determine the percentage of reinforcement the column should have and the X value. For example, if 3% reinforcement was chosen, we would use N/21.**

Column area (Ac) can be estimated by

Reinforcement Percentages for high yield steel | X in N/X |
---|---|

1% | 15 |

2% | 18 |

3% | 21 |

The maximum amount of reinforcement in concrete members (beams, columns, or slabs) should not exceed 4%.

Ac_req = N/X

where X is value given in table above

To estimate the applied moment on the columns, it is suggested to multiply the axial load from the floor above the column by:

le = β x l

where l = full length

β = Values from table below

End condition 1 = column end is fully restrained by moment connection

End condition 2 = column end is partially restrained by monolithic connection

End condition 3= column end is simply supported

**7. Determine the required concrete area**Ac_req = N/X

where X is value given in table above

**8. Determine the dimensions of the concrete column that has dimensions, b and h, which would give Ac_prov = b x h > Ac_req****9, Determine applied moment on the columns**To estimate the applied moment on the columns, it is suggested to multiply the axial load from the floor above the column by:

- 25 – interior columns
- 5 – edge columns
- 2 – corner columns

**Detail Design****1. Find the effective height, le, of the column**le = β x l

where l = full length

β = Values from table below

End condition 1 = column end is fully restrained by moment connection

End condition 2 = column end is partially restrained by monolithic connection

End condition 3= column end is simply supported

**2. Determine whether the column is a short column.**

If ley / b < 15 and lex / h < 15, it is a short column.

If both ratios are larger than 15, it is a slender column.

where lex = effective height in respect of the major axis

ley = effective height in respect of the minor axis,

Normally, reinforced columns should be designed as short, not slender.

**3. Find required area of steel reinforcement, Asc_req**

Sufficient steel reinforcement content and reinforcement placement help to resist cracking in the concrete column. Additional reinforcement should be used, such as binders, vertical links, or ties. These additional reinforcement resist lateral buckling induced by compressive stresses of main reinforcement. A tie should be placed for every corner bar. The distance from one reinforcement bar to another should be no less than 150mm.

Reinforcement near the concrete surface are more effective at resisting bending moment forces than reinforcement placed at the centre of the column.

Equation for a short and braced column which supports roughly symmetrical arrangement of beams and where these beam properties and sizes do not differ by more than 15% is shown below.

**N = 0.35 x fcu x Ac + 0.67 x Asc_req x fy**

Where fcu = characteristic strength of concrete (N/mm^2)

Ac = area of concrete (mm^2)

Fy = yield strength of reinforcement (N/mm^2)

Asc = area of reinforcement

Note: If Asc_req is negative, use the equation below.

Asc_req = 0.4% x Ac_nominal

Note: The design moment for slender columns includes an additional moment induced by eccentricity of the geometry section.

**4. Find a suitable number of reinforcement bars and the size of the reinforcement bars, ______ T ______**

**5. Find the area provided by the reinforcement bars designed, As_provc**

## Reinforced Concrete: Beam Design

**Scheme Design**

The applied loads include direct compression forces, as well as, compressive and tensile stresses that are caused by sagging bending moments to the beam. The induced compressive stresses are located in the material fibres above the neutral axis of the member and the induced tensile stresses are located below the neutral axis.

**1. Determine fy and fcu according to required material properties**

**2. Determine preliminary dimensions of the beam, b and h**

**3.**

**Find effective depth, d**

d = h - cover - bar diameter

Concrete covers are to be designed for requirements of fire resistance and durability.

**4. Find span/depth ratio, L/d and make sure that L/d is less than 20**

Deflection should be checked using the span/depth ratio.

Cracking should be designed for SLS and should meet the requirements of minimum reinforcement needed and spacing.

**Detail Design**

**1. Find w**

w = 1.4DL + 1.6LL

**2. Find the design moment and shear, M and V**

Simply supported with Uniformly Distributed Load

Simply supported with Concentrated Load

Cantilever Beam with Uniformly Distributed Load

Fixed ends with Uniformly Distributed Load

Fixed Ends with Concentrated Load at the Center

t

The effective span of beams, l, should be assumed to be the effective span of the member in its simply supported condition for conservative purposes. This span equals the exact distance between supports.

Depth to neutral axis, x, from compression face is limited to:

x ≤ 0.5d for fcu ≤ 45 N/mm^2

x ≤ 0.4d for 45 < fcu ≤ 70 N/mm^2

x ≤ 0.33d for 70 < fcu ≤ 100 N/mm^2

The design ultimate moment M should be designed greater than the ultimate bending moment.

*Also applicable to flanged beams when the neutral axis of the beam lies within the flange

The effective span of beams, l, should be assumed to be the effective span of the member in its simply supported condition for conservative purposes. This span equals the exact distance between supports.

**2. Establish concrete grade, fcu, in N/mm^2****3. Find the depth to neutral axis, x, in mm**Depth to neutral axis, x, from compression face is limited to:

x ≤ 0.5d for fcu ≤ 45 N/mm^2

x ≤ 0.4d for 45 < fcu ≤ 70 N/mm^2

x ≤ 0.33d for 70 < fcu ≤ 100 N/mm^2

**3. Design rectangular beams for flexure**

The design ultimate moment M should be designed greater than the ultimate bending moment.

*Also applicable to flanged beams when the neutral axis of the beam lies within the flange

K' = 0.156 for fcu ≤ 45 N/mm^2

K' = 0.120 for 45 < fcu ≤ 70 N/mm^2

K' = 0.094 for 70 < fcu ≤ 100 N/mm^2

If K ≤ K', compression reinforcement not required

K' = 0.120 for 45 < fcu ≤ 70 N/mm^2

K' = 0.094 for 70 < fcu ≤ 100 N/mm^2

If K ≤ K', compression reinforcement not required

x = (d-z)/0.45 for fcu ≤ 45 N/mm^2

x = (d-z)/0.40 for 45 < fcu ≤ 70 N/mm^2

x = (d-z)/0.36 for 70 < fcu ≤ 100 N/mm^2

x = (d-z)/0.40 for 45 < fcu ≤ 70 N/mm^2

x = (d-z)/0.36 for 70 < fcu ≤ 100 N/mm^2

If K > K', compression reinforcement required

x = (d-z)/0.45 for fcu ≤ 45 N/mm^2

x = (d-z)/0.40 for 45 < fcu ≤ 70 N/mm^2

x = (d-z)/0.36 for 70 < fcu ≤ 100 N/mm^2

x = (d-z)/0.45 for fcu ≤ 45 N/mm^2

x = (d-z)/0.40 for 45 < fcu ≤ 70 N/mm^2

x = (d-z)/0.36 for 70 < fcu ≤ 100 N/mm^2

The maximum amount of reinforcement in concrete members (beams, columns, or slabs) should not exceed 4%.

Shear stress in beams

Usually, the shear force and the shear stress should be obtained from the face of the support.

**5. Design rectangular beams for shear**Shear stress in beams

Usually, the shear force and the shear stress should be obtained from the face of the support.

Shear reinforcement

Shear reinforcement should be designed for ULS and should be provided in the form of vertical links or bent-up bars. Shear forces are transferred to the vertical links that act with diagonal concrete struts in compression. Therefore, in beams, the links will act in tension and the concrete in compression.

Shear reinforcement are required to resist the following failure mode caused by shear:

a. If v < 0.5vc, minimum links should be provided.

b. If 0.5vc < v < vc + vr, links should be provided, in which the area of shear reinforcement provided is

Shear reinforcement should be designed for ULS and should be provided in the form of vertical links or bent-up bars. Shear forces are transferred to the vertical links that act with diagonal concrete struts in compression. Therefore, in beams, the links will act in tension and the concrete in compression.

Shear reinforcement are required to resist the following failure mode caused by shear:

- Inclined tensile cracks on beam
- Inclined tensile stress failure caused by shear

a. If v < 0.5vc, minimum links should be provided.

b. If 0.5vc < v < vc + vr, links should be provided, in which the area of shear reinforcement provided is

and where fcu not reater than 80.

c. If vc +vr < v < 0.8√fcu or v = 7 N/mm², links or links with bent-up bars should be provided. Links should not be more than 50% of shear resistance.

c. If vc +vr < v < 0.8√fcu or v = 7 N/mm², links or links with bent-up bars should be provided. Links should not be more than 50% of shear resistance.

Concrete shear stress, vc

**5. Determine whether maximum deflection is below deflection capacity**

Allowable Limit = L/250

## Reinforced Concrete: Slab Design

Types of suspended slabs that are considered (slabs which are supported by beams, columns, or walls)

No slabs should be less than 125 mm thick due to fire resistance requirements.

Two way spanning slabs can be 90% of thickness of one way spanning slabs

w = 1.4DL + 1.6LL

Find M and V equations above (Refer to beam calculations).

Find K and z

- Solid slabs
- These slabs are made out of solid concrete with reinforcement that resist tension. Slabs can be in-situ or profiled metal decking. The top reinforcement can be a steel mesh for the use of fire resistance. The bottom reinforcement can be a metal decking for tension reinforcement purposes.

- Ribbed slabs
- These slabs can achieve the same structural strength than solid slabs with less concrete required. Ribbed slabs can be series of in-situ concrete ribs that are cast monolithically with voids caused by removable formers. Ribbed slabs can also be a hollow slabs with permanent void formers.

- Flat slabs
- These slabs with flat soffits do not require the support of beams. Drops are often used to form a thick stiffening part between the columns and the slab.

- Waffle slabs
- These slabs are solid and flat with void formers in the soffits. There are series of 1m wide concrete beams that can be designed for moment bending.

No slabs should be less than 125 mm thick due to fire resistance requirements.

Two way spanning slabs can be 90% of thickness of one way spanning slabs

**1. Find w**w = 1.4DL + 1.6LL

**2. Find the design moment and shear, M and V**Find M and V equations above (Refer to beam calculations).

**3. Design slab for flexure using One Way Slab method**Find K and z

Find percentage of reinforcement in concrete area (Ast/bd = %)

Reinforcement bars should be designed to fulfill the minimum area capacity and should be constructed in both directions in the slab. Steel reinforcements help to resist cracking and to distribute concentrated loads throughout the slab.

Reinforcement bars should be designed to fulfill the minimum area capacity and should be constructed in both directions in the slab. Steel reinforcements help to resist cracking and to distribute concentrated loads throughout the slab.

The maximum amount of reinforcement in concrete members (beams, columns, or slabs) should not exceed 4%.

Refer to here for rules for each constant in the concrete shear stress equation below.

**4. Find the number of reinforcement bars and the size of the reinforcement bars, ____ T ______.****5. Find Asprov.****6. Design slab for shear.**Refer to here for rules for each constant in the concrete shear stress equation below.

Minimum steel required = 0.13%

Punching shear forces (shear forces around the perimeter of columns) are usually the critical design case for flat slab foundations. Effective shear is the shear force that takes moment forces caused between the slab and the column and the shear force over the area supported by the column.

Punching shear checks in flat slabs

**7. Check punching shear**Punching shear forces (shear forces around the perimeter of columns) are usually the critical design case for flat slab foundations. Effective shear is the shear force that takes moment forces caused between the slab and the column and the shear force over the area supported by the column.

- Effective shears
- Internal columns -> Veff = 1.15V
- Corner columns -> Veff = 1.25V
- Edge columns -> Veff = 1.4V

Punching shear checks in flat slabs

- Shear stress, vo = Veff/ Uod < 0.8sqrt(Fcu) or 7 N/mm^2
- Uo is the column perimeter that touches the slab

- Shear forces should be checked at certain perimeters of slab enveloping the column. Shear forces should be checked starting at the first perimeter of 1.5d around the column face. Then, shear forces should be checked subsequently at perimeters of 0.75d intervals.